Question

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$a^2{b}^2{x}^2-\left(4b^4-3a^4\right)x-12a^2{b}^2=0$, $a\ne 0$ and $b\ne 0$ [CBSE 2006]

Answer 1

The given equation is $a^2{b}^2{x}^2-\left(4b^4-3a^4\right)x-12a^2{b}^2=0$.

Comparing it with $A{x}^2+Bx+C=0$, we get

A = $a^2{b}^2$, B = $-\left(4b^4-3a^4\right)$ and C = $-12a^2{b}^2$

∴ Discriminant, D = $B^2-4AC={\left[-\left(4b^4-3a^4\right)\right]}^2-4\times a^2{b}^2\times \left(-12a^2{b}^2\right)=16b^8-24a^4{b}^4+9a^8+48a^4{b}^4=16b^8+24a^4{b}^4+9a^8={\left(4b^4+3a^4\right)}^2>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(4b^4+3a^4\right)}^2}=4b^4+3a^4$

$$\begin{gathered} \therefore \alpha =\frac{-B+\sqrt{D}}{2A}=\frac{-\left[-\left(4b^4-3a^4\right)\right]+\left(4b^4+3a^4\right)}{2\times a^2{b}^2}=\frac{8b^4}{2a^2{b}^2}=\frac{4b^2}{a^2} \\ \beta =\frac{-B-\sqrt{D}}{2A}==\frac{-\left[-\left(4b^4-3a^4\right)\right]-\left(4b^4+3a^4\right)}{2\times a^2{b}^2}=\frac{-6a^4}{2a^2{b}^2}=-\frac{3a^2}{b^2} \end{gathered}$$

Hence, $\frac{4b^2}{a^2}$ and $-\frac{3a^2}{b^2}$ are the roots of the given equation.